4 real periods
tommy1729
with
F(x) = A + B + C + D
where A , B , C , D have 4 real periods and satisfy R -> R
A = a_0 + a_1 sin(P_1 x) + a_2 sin(2 * P_1 x) + a_3 ...
+ A_1 cos(P_1 x) + A_2 cos(2 * P_1 x) + A_3 ...
B = b_0 + b_1 sin(P_2 x) + b_2 sin(2 * P_2 x) + b_3 ...
+ B_1 cos(P_2 x) + B_2 cos(2 * P_2 x) + B_3 ...
ETC ( real fourier series )
1) given a function G(x) R -> R, how do we decide if it belongs to this set ?
2) how do we find the 4 real periods P_1, P_2, P_3 and P_4 ?
3) how do we find the coefficients of A B C and D ? (*)
(*) knowing the answer to 2 questions also solves the 3rd.
(note: of course G(x) is not given as the sum of 4 fourier series nor are its periods given in advance)
regards
tommy1729
tommy1729
well ? where are the calculus experts now ?
neilist
>
> - Show quoted text -
You are just starving for attention, Tommy. The calculus experts
don't answer you, and you get impatient.
Tsk tsk.
Hey, Pat Buchanan, that great American (snicker and guffaw) predicted
last week that Belgium is going to be the next European nation to
split up. What, Belgium is in worse shape than Serbia which might
break off Kosovo? Who would have known?
Is it because you, Tommy, troll the streets, causing civil unrest
there? Spewing your "math" and demanding a response from your fellow
Belgians?
You are indeed the James Harris of Belgium!
Alas for that poor nation.
;-)
tommy1729
maybe there are no calculus experts here
why dont you do the math ???
LOL
>
> Tsk tsk.
>
> Hey, Pat Buchanan, that great American (snicker and
> guffaw)
i dont know who that is...
and since you have a big intrest in people like JSH , Tom potter and others i dont even wanna know who that is
predicted
> last week that Belgium is going to be the next
> European nation to
> split up. What, Belgium is in worse shape than
> Serbia which might
> break off Kosovo? Who would have known?
it might amaze you but
1) the guy may be right, and therefore may be smarter than you think
2) we dont speak german alot but rather dutch , you dont know the difference
3) belgium is in bad shape because we are together, at least thats what some politicians believe.
the upper half thinks it is better off without the lower part.
well at least some politicians do.
4) there is no war or economic crisis , only political crisis
5) dont you dare to insult my country or people or i will spam you forever you asshole
6) you dont know shit about belgium politics
7)comparing belgium to Serbia is like comparing the french to the russians or the italians to the brittish.
it is BOGUS.
not that that surprises me from you.
8) buy a belgium newspaper , it not just that guy who says it , there are politician in belgium itself who claim a split.
9) at least we dont have an " iraq policy "
10) or listen to britney spears or paris hilton like you americans like to do.
11) we are not impressed by your superstars like paris hilton , pete doherty , pink and those other drug addicts.
12) dont speak about things you know nothing about
13) where is your math in this post
14) your an asshole , look at your own country and president !! you should be asshamed with your " iraq policy " , not to mention other stuff.
15) our main city in belgium is an important place for meetings of the european union for instance.
so dont compare to serbia or claim other bogus.
>
> Is it because you, Tommy, troll the streets, causing
> civil unrest
> there?
oh really funny neilist.
its easier to say that , than to understand the politics of another country hmm.
certainly if you confuse their language with german.
buy a newspaper !
Spewing your "math" and demanding a response
> from your fellow
> Belgians?
>
> You are indeed the James Harris of Belgium!
you are an idiot, you know nothing about politics of other countries.
>
> Alas for that poor nation.
>
> ;-)
>
you are PWNED neilist.
tommy1729
neilist
<snip>
> > Hey, Pat Buchanan, that great American (snicker and
> > guffaw)
>
> i dont know who that is...
Well then, I guess you don't know the politics of another country,
either. Pat Buchanan WAS a U.S. presidential candidate. And he got a
few votes, too. Stole them from Gore in Florida, and thus we have the
current U.S. president.
> 6) you dont know shit about belgium politics
> 7)comparing belgium to Serbia is like comparing the french to the russians or the italians to the brittish.
> it is BOGUS.
Well, I don't hear of France or Italy breaking up. Russia already
did, and might again. Great Britain is already giving greater
autonomy to Scotland and Wales.
Yet you have Belgium and Serbia both mentioned in the media within a
week of each possibly breaking up. That is very comparable. Serbia
is understandable - they already lost Montenegro, and now Kosovo
possibly.
But Belgium? You'd think that developed established country in the
heart of Western Europe would be more stable than, say, Italy. I
guess not!
hahahahahahahahahahahahahahahahaha (Italy) hahahahahahahahahahahahaha
> 15) our main city in belgium is an important place for meetings of the european union for instance.
And yet you can't keep your poor nation together. Europe
consolidates, yet Belgium splits apart. How ironic!
> so dont compare to serbia or claim other bogus.
But both might break up! That's definitely comparable.
And if Belgium breaks up before Serbia? Won't that hurt your pride!
> you are an idiot, you know nothing about politics of other countries.
And you don't know about that great American stateman Pat Buchanan.
For shame! (snicker)
Perhaps you never heard of that socialist American political candidate
James Harris, either?
I should say, that LOSER James Harris, candidate of insanity! Yeah
you socialists, you have a James Harris as YOUR candidate!
hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha
With James Harris as a candidate, how can you lose? He's got his
mighty Hammer and already has the ARMY!
hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha
David C. Ullrich
wrote:
The calculus experts are pretty certain that since
you insist on spouting nonsense about utterly elementary
topics in complex analysis there's very little point
in trying to explain the theory of "almost-periodic
functions" to you.
************************
David C. Ullrich
tommy1729
dont speak for others David.
what you really want to say is you dont understand it either hmm
spouting nonsense like integral x dx = 2x + 1
oh no , wait thats your equation.
you are a very strange "prof" David.
1) you wrote integral x dx = 2x + 1 for example (there are similar "math demonstrations" )
but hey lets disregard that ,
2) when the questions become hard , you ignore the thread or you say it is trivial...
3) ... without solving it !!!
seems like you cant do the math , or you are not very good at helping people with it.
you even refuse to give a decent answer.
if you cant do math or you refuse to help people with math , that makes you a very "special type" of prof.
even if i only assume the second.
but whats even worse:
you have no idea what this is all about dont you ?
if the question is so simple , why dont you give the formula for the 4 periods ?
you refer to " almost periodic functions"
if you truely are a prof , i assume a prof in physics.
why ? because only in physics this could work out.
in math , this is a mistake.
it works out for problems like n-body problems or quasicrystals, because they can be solved with the math of almost periodic functions ...
because they ARE almost periodic functions
however my questions was quite different.
definition of almost periodic function :
F(x) is almost-periodic if every sequence of translates of F(x) has a uniformly convergeant subsequence.
almost periodicity is a property of dynamical systems that appear to retrace their paths through phase space, but not exactly.
if we consider functions based on time then a theorem of Kronecker ( yes kronecker THE ENEMY OF CANTOR !! doing math cantor could not ! ) from diophantine approximation can be utilized to show that a particular configuration that ocurs once, will recur to within any specified accuracy in finite time.
(haha you yourself lead me to a theorem of kronecker while you are a cantorian :p)
the problems are that this does not apply well with my questions because of at least 2 reasons :
1) for almost periodic functions we have the condition
ABS [F(t) - F(t + P)] < B
where abs is the absolute value , t is time and the real parameter of the function , P is the almost period and B is finite real number.
compare with my question ; 1) the restriction B is not present , my fourrier series can go to infinity !!!
so the condition of almost periodic functions is not matched !!
(it is usually so in physics because we have finite distance as in n-body problems and similars )
so B is not matched , and neiter is P as shown in 2)
2) since we have 4 periods we have a lot of periods ; therefore none of them can be dominant so we dont have an approximation to a certain period --> P cannot be determined consistantly.
3) well ill spare you that , not to embarras you to much :p
perhaps you where thinking what about quasiperiodic functions then ??
1) not invented by quasi despite he understands them
2) dont work either
i guess you will have to invent something new to analyse
4-periodic functions
perhaps ullrich-almost-3-periodic-functions
or ullrich-almost-5-periodic-functions
tommy1729
"tommy is tommy, crazy but crazy like a fox"
"polysigned is the future"
"what is g(g(x))=f(x) to easy for you ? "
"math is bigger than physics"
"tommy is undecidable"
>
>
> ************************
>
> David C. Ullrich
David C. Ullrich
wrote:
Once again, you talk about mathematics that you don't
have any understanding of. You should really learn to
stop doing that. The sum of two (continuous) periodic
functions is almost periodic. That's a fact, even
if it's not mentioned in the place where you found
the (garbled) definition above.
>perhaps ullrich-almost-3-periodic-functions
>or ullrich-almost-5-periodic-functions
>
>tommy1729
>"tommy is tommy, crazy but crazy like a fox"
>"polysigned is the future"
>"what is g(g(x))=f(x) to easy for you ? "
>"math is bigger than physics"
>"tommy is undecidable"
>
>
>>
>>
>> ************************
>>
>> David C. Ullrich
************************
David C. Ullrich
tommy1729
now i showed you didnt :)
You should really learn to
> stop doing that. The sum of two (continuous) periodic
> functions is almost periodic.
i agree with that :-)
That's a fact, even
> if it's not mentioned in the place where you found
> the (garbled) definition above.
lol, i agree with you :)
read the title david :
FOUR periods
not just 2 ...
you know 2 + 2 = 4 so just adding 2 periods is insufficient.
why cant you just admit you are wrong.
or give the solution
you never do either.
because you cant solve it , and neiter admit it.
but hey look on the bright side i agree on what you said ....
yet it is not relevant of course
you really have to do better than this if you want to give the solution or show that i am wrong.
answering a 4 period question with a 2 period answer ...
lol
perhaps you want to add 2 almost periodic functions and call them ullrich-almost-5-periodic-functions to tackle the problem...
> and it works fine with the above definition.
>
> >perhaps ullrich-almost-3-periodic-functions
> >or ullrich-almost-5-periodic-functions
> >
> >tommy1729
> >"tommy is tommy, crazy but crazy like a fox"
> >"polysigned is the future"
> >"what is g(g(x))=f(x) to easy for you ? "
> >"math is bigger than physics"
> >"tommy is undecidable"
> >
> >
> >>
> >>
> >> ************************
> >>
> >> David C. Ullrich
>
>
> ************************
>
> David C. Ullrich
regards funny david
tommy1729
hagman
Maybe that's why Belgium was on sale at ebay...
tommy1729
lol :)
but hey what happended to the claim of neilist :
"i am certain tommy is not from belgium and is not dutch"
there was going to be a proof for that too :)
dont waste your time on neilist hagman , you have math talent , whereas neilist never does math.
regards
tommy1729
hagman
> david wrote:
> > On Thu, 20 Sep 2007 16:54:44 EDT, tommy1729
> > wrote:
>
> > >David C Ullrich wrote
>
> > >> On Wed, 19 Sep 2007 09:35:05 EDT, tommy1729
OK, I'll read the rest of that wikipedia article as well.
>
> > >if we consider functions based on time then a
> > theorem of Kronecker ( yes kronecker THE ENEMY OF
> > CANTOR !! doing math cantor could not ! ) from
> > diophantine approximation can be utilized to show
> > that a particular configuration that ocurs once, will
> > recur to within any specified accuracy in finite
> > time.
>
> > >(haha you yourself lead me to a theorem of kronecker
> > while you are a cantorian :p)
>
> > >the problems are that this does not apply well with
> > my questions because of at least 2 reasons :
>
> > >1) for almost periodic functions we have the
> > condition
>
> > >ABS [F(t) - F(t + P)] < B
>
> > >where abs is the absolute value , t is time and the
> > real parameter of the function , P is the almost
> > period and B is finite real number.
Apparently you didn't read the article thotoughly enough.
With your definition I could take F=characteristic function of
rationals,
B=1.1, P=17 and qould obtain that F is almost-periodic with
almost period 17.
Instead, the definition should be that for every B>0 you
can find a P>0 (thus depending on B) such that etc.
Assume A has the property
forall eps>0: exists P>0: forall x: |A(x)-A(x+P)| < eps
(aka. almost periodic)
and B is continuous and periodic with period Q.
Then C:=A+B is almost periodic.
Proof:
Assume eps>0 given.
Let eps_n be a sequence of positive numbers converging to 0.
For each n, there is an eps_n-almost period P_n of A.
The sequence P_n mod Q has an accumulation point R.
Wlog. P_n mod Q converges to R.
Let k_m be a sequence of positive integers such that
k_m*R mod Q converges to 0 (or Q if you like).
For each m, we have eps_n < eps/(2*k_m) for n big enough.
By taking a subsequence of the P_n, we may assume wlog. that
eps_n < eps/(2*k_n) for all n.
Again by taking a subsequence of the P_n (but not the k_n),
we may assume that k_n*P_n mod Q converges to 0 (i.e. the
difference between k_n*P_n and the closest integer multiple of Q
converges to 0).
Observe that k_n*P_n is an eps/2-almost period of A
per telescope summing.
Next, the Q-periodic functions B_n(x):=B(x)-B(x+k_n*P_n)
converge pointwise to 0.
The B_n are clearly equicontinuous and uniformly bounded.
Hence, by Arzela-Ascoli, some subsequence converges uniformly
(of course also towards the pointwise limit)
Wlog. B_n -> 0 uniformly.
Thus, for n big enough, |B_n(x)| < eps/2 for all x.
|C(x) - C(x+k_n*P_n)| <= |A(x)-A(x+k_n*P_n)| + |B(x)-B(x+k_n*P_n)|
< eps/2 + eps/2 = eps for all x.
Thus C is almost periodic. QED
(Gee, I'm so glad eps/2 worked out. I hate it when you
note that you have to go back and replace all eps/2 by eps/3) ;)
Corollary: The sum for finitely many continuous periodic functions
is almost periodic.
Proof: trivial induction.
I'm sorry that my proof is so lengthy ansd unelegant, but
I have read this thread and especially the definition of almost
periodic just 10 minutes ago and I'm rather algebraist than analyst.
Someone who has considered the original problem
thoroughly for some time and can tell immediately when D. Ullrich
is wrong, could surely make that proof more stringent.
neilist
>
> - Show quoted text -
Eat my reposts, asshole Tommy.
tommy1729
lol
continuous :)
where did i state that restriction in the original post :)
nice try hagman
better math than david
respect , but still not good enough to beat tommy :)
or to answer the questions from the OP.
>
> I'm sorry that my proof is so lengthy ansd unelegant,
> but
> I have read this thread and especially the definition
> of almost
> periodic just 10 minutes ago and I'm rather
> algebraist than analyst.
thats ok
> Someone who has considered the original problem
> thoroughly for some time and can tell immediately
> when D. Ullrich
> is wrong
i can tell you that :)
, could surely make that proof more
> stringent.
continuous :)
i never restricted to that :)
even with the help of a real mathematician like hagman , david's ideas cannot be defended :)
at least he doesnt have an additude of
i dont know and wont admit , and if i did know , i refuse to help you and call it trivial or insult you.
so regards
tommy1729
David C. Ullrich
wrote:
You really have no idea how funny it is for _you_ to complain about
people being insulting?
No, you never said the functions you were talking about were
continuous. You just said
>A = a_0 + a_1 sin(P_1 x) + a_2 sin(2 * P_1 x) + a_3 ...
> + A_1 cos(P_1 x) + A_2 cos(2 * P_1 x) + A_3 ... ,
which is more or less meaningless without some comment
on what sort of function you're talking about and/or
what sort of convergence you have in mind. I assumed you
were talking about continuous functions, that being sort
of the simplest case. If you explain what you did have
in mind there will be a version of almost-periodicity
that handles that case.
>so regards
>tommy1729
************************
David C. Ullrich
hagman
Hm, in fcat you did not.
It seems I took that from David's post as a nice to have condition.
However,
>A = a_0 + a_1 sin(P_1 x) + a_2 sin(2 * P_1 x) + a_3 ...
> + A_1 cos(P_1 x) + A_2 cos(2 * P_1 x) + A_3 ...
may imply continuos depending on what type of convergence
you want to imply.
Or should the functions be merely L^2?
>
> nice try hagman
>
> better math than david
>
> respect , but still not good enough to beat tommy :)
My math may be closer to tommy than David, but
my math is probably not better than David's.
tommy1729
AHA , you admit for once :-)
You just said
>
> >A = a_0 + a_1 sin(P_1 x) + a_2 sin(2 * P_1 x) + a_3
> ...
> > + A_1 cos(P_1 x) + A_2 cos(2 * P_1 x) + A_3
> ... ,
>
> which is more or less meaningless without some
> comment
no it is more or less not restriction too continuous...
it is more "general".
not meaningless.
there you insulting me again ...
claiming i posted meaningless stuff.
yet you admitted you wrongly restricted to continuous and are beginning to understand the mistake in your reply.
> on what sort of function you're talking about and/or
> what sort of convergence you have in mind. I assumed
> you
i was talking about the functions i defined of course ...
in general a questions refers to the definitions given before it.
:-)
> were talking about continuous functions, that being
> sort
> of the simplest case.
you mean : the simplest approach you had.
way to simple.
If you explain what you did
> have
> in mind there will be a version of almost-periodicity
> that handles that case.
explained what i had in mind ????
the functions where DEFINED !!!
if you start having fantasies about things i did not say , thats your responsibility , not mine.
everything was CLEARLY defined by the series.
>
> >so regards
> >tommy1729
>
>
> ************************
>
> David C. Ullrich
well at least your making progress by seeing your mistake
so
regards
tommy1729
David C. Ullrich
wrote:
That doesn't answer the question.
>in general a questions refers to the definitions given before it.
>
>:-)
>
>
>> were talking about continuous functions, that being
>> sort
>> of the simplest case.
>
>you mean : the simplest approach you had.
>
>way to simple.
>
>
> If you explain what you did
>> have
>> in mind there will be a version of almost-periodicity
>> that handles that case.
>
>explained what i had in mind ????
>
>the functions where DEFINED !!!
>
>if you start having fantasies about things i did not say , thats your responsibility , not mine.
>
>everything was CLEARLY defined by the series.
>
>
>>
>> >so regards
>> >tommy1729
>>
>>
>> ************************
>>
>> David C. Ullrich
>
>well at least your making progress by seeing your mistake
>
>so
>
>regards
>tommy1729
************************
David C. Ullrich